Given the equation: $ x^2 + \ln^2x = n $.

Let $x(n)$ - root of the equation and $x(n) > 1$

The problem: find missing terms in following sum: $$ x(n) \approx ... + O(\frac{\ln^4n}{n})$$

I've found the first term and get next result:

$$ x(n) \approx \sqrt{n} + ... + O(\frac{\ln^4n}{n}) $$

How to determine remaining terms of the sum?


Well, let $x=\sqrt{n}-y$, where $y=o(\sqrt n)$. Now: $$(\sqrt{n}-y)^2+\ln^2(\sqrt{n}-y)=n$$ $$-2\sqrt{n}\cdot y + y^2 + \ln^2\sqrt n + \left(\ln(1-\frac{y}{\sqrt n})\right)^2=0$$ Now discard the smaller terms and see what remains: $$-2\sqrt{n}\cdot y + \ln^2\sqrt n = 0$$ $$y = \left.\left({1\over2}\ln n\right)^2\right/2\sqrt{n}$$ (plus some small-o, of course).

  • 2
    $\begingroup$ Can you explain why $y^2=o(n)$ has been discarded? It does not look too small. $\endgroup$ – A.Γ. Oct 16 '15 at 7:07
  • $\begingroup$ If $y=o(\sqrt n)$, then $y^2=o(y\cdot\sqrt n)$. $\endgroup$ – Ivan Neretin Oct 16 '15 at 7:27
  • $\begingroup$ Yes, but $y=o(\sqrt{n})$ could make $y^2$ much larger than $\ln^2\sqrt{n}$. Thus one cannot discard the former and keep the larger as the answer is presently doing. More work is needed. $\endgroup$ – Did Oct 16 '15 at 7:29
  • 1
    $\begingroup$ If $y^2$ were as large as $\ln^2n$ or larger, then $y\sqrt n$ would be even larger, and wouldn't have anything to cancel out with. I agree, of course, that my answer is but a sketch rather than a proof. $\endgroup$ – Ivan Neretin Oct 16 '15 at 7:41
  • 1
    $\begingroup$ In your example we don't have the reason to believe that the quadratic term is small; $\epsilon$ may be small, but we don't know what $y$ is. In this problem, we do know that $y=o(\sqrt n)$; that was our assumption from the very beginning. Anyway, if you insist, let's just discard the second logarithm term (which is $o(1)$ anyway), solve the quadratic equation for $y$, approximate the solution and get exactly the same result. $\endgroup$ – Ivan Neretin Oct 16 '15 at 7:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.